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Find a real number $x$ such that $x^5 − x − 1 = 0$.

I have already proven that such a number exists, I now try to find which number it is.

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  • $\begingroup$ Iteration tells me its approximate value is $1.167303978$ (I iterated $x_{n+1}=\sqrt[5]{x_n+1}, x_1=1$) $\endgroup$ Jun 7, 2018 at 11:53
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    $\begingroup$ A real polynomial with odd degree has at least a real root. $\endgroup$
    – Cesareo
    Jun 7, 2018 at 11:55
  • $\begingroup$ I'm voting to close this question because it's essentially answered already: you know there is a root. The comments tell you how to find a numerical approximation. There is no way to express that root with an expression involving roots and other algebra. $\endgroup$ Jun 7, 2018 at 11:58
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    $\begingroup$ You can write the root as a nice-looking series $\endgroup$
    – user567668
    Jun 7, 2018 at 12:03
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    $\begingroup$ Do you know the notion of "Galois group"? That is the concept to use to answer whether or not the solution can be expressed in terms of radicals. $\endgroup$
    – GEdgar
    Jun 7, 2018 at 13:15

2 Answers 2

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You can use numerical methods such as Newtons method .

Define $h(x) = x^5-x-1$

$h'(x) = 5x^4-1$

$x_{n+1} = x_n-\dfrac{h(x_n)}{h'(x_n)}$

$x_1 = x_0 - \dfrac{h(x_0)}{h'(x_0)}$

With $x_0 = 1$ we get the following;

$\begin{pmatrix}n&&&&x_n\\0&&&&1\\1&&&&1.25\\2&&&&1.178459\\3&&&&1.167537\\4&&&&1.167304\\\vdots\end{pmatrix}$

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    $\begingroup$ Right, so this defines a Cauchy sequence $x_n$ that converges to a real number that is the solution of $x^5-x-1$. If we take equivalence class of this Cauchy sequence, we have now successfully constructed the real number we were looking for. $\endgroup$
    – Kasper
    Jun 7, 2018 at 12:08
  • $\begingroup$ yes , thats correct. $\endgroup$ Jun 7, 2018 at 12:09
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You can use an iteration to find its approximate value:

Rearrange for:$$x^5=x+1\to x=\sqrt[5]{x+1}$$ Then iterate: $$x_{n+1}=\sqrt[5]{x_n+1}, x_1=1$$

Calculator accuracy gives me $x\approx1.167303978$

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